Answer to Question #109587 in Calculus for Alamin

Question

Find the limits.
f(x) = 2x2 − 3x + 15, g(x) =
3

x + 3

(a)

lim
x→3
f(x) =


(b)

lim
x→24
g(x) =


(c)

lim
x→3
g(f(x)) =

Answer 1

$(a) \lim_{x\to 3}f(x) = \lim_{x\to 3} (2x^2-3x+15)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \lim_{x\to 3} (2(3)^2-3(3)+15)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 24\\ (b) \lim_{x\to 24}g(x) = \lim_{x\to 24} \frac{3}{x+3}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \lim_{x\to 24}\frac{3}{24+3} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{9}\\ (c) \lim_{x\to 3}g(f(x)) = \lim_{x\to 3} \frac{3}{f(x)+3}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \lim_{x\to 3}\frac{3}{2x^2-3x+18} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{9}\\$